Class 10 Mathematics Quadratic Equations Extra Questions
Chapter 4: Quadratic Equations
Topic–1: Quadratic Equations
Very Short Answer Type Questions: [1 Mark]
1. If 8 is the root of the equation x2 – 10x + k = 0, then find the value of k.
Sol. (8)2 – 10 × 8 + k = 0
⇒ 64 – 80 = –k ⇒ k = 16
2. Is x = 3 a root of the equation 2x2 – 5x – 3 = 0?
Sol. 2(3)2 – 5(3) – 3 = 0
⇒ 18 – 15 – 3 = 0
So, x = 3 is the root of 2x2 – 5x – 3 = 0.
3. (x + 2)3 = x(x2 – 1) is a quadratic equation. Is it true?
Sol. (x + 2)3 = x(x2 – 1)
⇒ x3 + 12x + 6x2 + 8 = x3 – x
⇒ 6x2 + 13x + 8 = 0
So, it is a quadratic equation.
4. Is x = 1 a root of the equation 3x2 – 2x – 1 = 0?
Sol. 3(1)2 – 2 × 1 – 1 = 3 – 2 – 1 = 3 – 3 = 0
So, x = 1 is a root of 3x2 – 2x – 1 = 0
5. For what value of k, x = 2 is a solution of kx2 + 2x – 3 = 0?
Sol. k(2)2 + 2 × 2 – 3 = 0
⇒ 4k + 4 – 3 = 0
k = − ¼
6. If x = 2 is a root of the equation 3×2 – 2kx + 5 = 0, then find k.
Sol. 3 × (2)2 – 2 × 2 × k + 5 = 0
⇒ 12 – 4k + 5 = 0 ⇒ k = 17/4
Short Answer Type Questions:
- Which of the following among the given expressions is a polynomial?
i) 1 + √x + x (ii) 1 + x2 (iii) x + 1/x
Sol. (i) The power of x in √x, that is in x 1/2 is not a non-negative integer.
∴ 1 + √x + x are not a polynomial.
(ii) Here, the power of x in every term of the given expression is non-negative.
∴ 1 + x2 is a polynomial.
(iii) The power of x in the term 1/x, that is in, x–1 is a negative integer ∴ x + 1/x is not a polynomial.
Ans.
is a polynomial; since the degree of x in each term of the given expression is a non-negative integer. The degree of this polynomial is 2 since the maximum power of x in the given expression is 2.
is not a polynomial, since in 1/x2, that is in x–2, the power of x is a negative integer.
Short Answer Type questions: [3 Marks]
1. Check whether the following equations are quadratic or not :
(i) (x – 1)(x – 3) = (x + 5)(x – 1)
(ii) x2 + 2x – 6 √x + 5 = 0
(iii) x3 – 4x2 + 1 = (x – 2)3
Sol. (i) (x – 1) (x – 3) = (x + 5) (x – 1)
⇒ x2 – 4x + 3 = x2 + 4x – 5
⇒ –4x – 4x + 3 + 5 = 0
⇒ –8x + 8 = 0
Which is not of the form ax2 + bx + c = 0.
Thus, (x – 1) (x – 3) = (x + 5) (x – 1) is not a quadratic equation.
(ii) x2 + 2x – 6 √x + 5 = 0
The given equation contains a term involving √x, i.e., x 1/2.
Since, x2 + 2x – 6 √x + 5 is not a quadratic polynomial, therefore,
x2 + 2x – 6 5 x + = 0 is not a quadratic equation.
(iii) x3 – 4x2 + 1 = (x – 2)3
⇒ x3 – 4x2 + 1 = x3 – 8 – 6x(x – 2)
⇒ –4x2 + 1 = –8 – 6x2 + 12x
⇒ –4x2 + 6x2 – 12x + 1 + 8 = 0
⇒ 2x2 – 12x + 9 = 0, which is a quadratic equation as it is of the form
ax2 + bx + c = 0. Thus, the given equation is a quadratic equation.
2. Determine whether the given values are solutions (roots) of the equation or not.
(ii) a2x2 – 3abx + 2b2= 0, x = a/b, x=b/a
⇒ 2 + 2 – 4 = 0
⇒ 4 – 4 = 0
⇒ 0 = 0, which is true.
x = √2 is a solution of the equation.
Now, putting x = −2 √2, we get
⇒ 8 – 4 – 4 = 0
⇒ 0 = 0, which is true.
So, x = −2 √2 is a solution of the equation.
(ii) a2x2 – 3abx + 2b2 = 0
Putting x = a/b in the given equation, we get
⇒ a4 – 3a2b2 + 2b4 = 0, which is not true.
x = a/b is not a solution to the given equation.
Again, by putting x = b/a in the given equation, we get
⇒ b2 – 3b2 + 2b2 = 0
⇒ –2b2 + 2b2 = 0
⇒ 0 = 0, which is true.
x = b/a is a solution of the given equation.
the following, find the value of k for which the given value is a solution of the given equation :
(i) x2 – x(a + b) + k = 0, x = a
(ii) kx2 + √2x – 4 = 0, x = 2
Sol. (i) Since x = a is a root of the given equation. Therefore, it satisfies the equation.
i.e., a2 – a (a + b) + k = 0
⇒ a2 – a2 – ab + k = 0
⇒ k = ab (ii) Since x = √2 is a root of the given equation. Therefore, it satisfies the equation.
⇒ b2 – 3b2 + 2b2 = 0
⇒ –2b2 + 2b2 = 0
⇒ 0 = 0, which is true.
x = b/a is a solution of the given equation.
3. In each of the following, find the value of k for which the given value is a solution of the given equation :
(i) x2 – x(a + b) + k = 0, x = a
(ii) kx2 + √2x – 4 = 0, x = 2
Sol. (i) Since x = a is a root of the given equation. Therefore, it satisfies the equation.
i.e., a2 – a (a + b) + k = 0
⇒ a2 – a2 – ab + k = 0
⇒ k = ab
(ii) Since x = √2 is a root of the given equation. Therefore, it satisfies the equation.
⇒ 2k + 2 – 4 = 0
⇒ 2k – 2 = 0 ⇒ k = 1
Long Answer Type questions: [4 Marks]
1. If x = – 2 and x = − 1/5 are solutions of the equation 5x2 + kx + λ = 0, find the values of k and λ.
Sol. Since x = –2 and x = − 1/5 are solutions of the equation, 5×2 + kx + λ = 0
∴ 5(–2)2 + k(–2) + λ = 0
⇒ 2k – λ = 20 …(i)
From (i), we get
λ = 2k − 20
Substituting this value of λ in (ii), we get
k − 5(2k − 20) = 1
⇒ k −10k +100 = 1
⇒ –9k = –99 ⇒ k = 11
∴ λ = 2k – 20 = 2(11) – 20 = 2
Hence, k = 11 and λ = 2
2. If x = 2/3 and x = –3 are the roots of the equation ax2 + 7x + b = 0, find the values of a and b. Sol. Since x = 2/3 is a root of ax2 + 7x + b = 0, we have,
⇒ 4a + 42 + 9b = 0
⇒ 4a + 9b + 42 = 0 …(i)
Again, x = –3 being a root of ax2 + 7x + b = 0, we have, a(–3)2 + 7(–3) + b = 0
⇒ 9a – 21 + b = 0
⇒ 9a + b – 21 = 0 …(ii)
Solving (i) and (ii), we get
a = 3 and b = – 6
3. If a and b are the roots of the quadratic equation x2 – p ( x + 1) – c = 0, then find the value of (a + 1) (b + 1).
Sol. We have x2 – p(x + 1) – c = 0
⇒ x2 – px – (p + c) = 0
⇒ Since a and b are roots of the equation
